Working through a pure maths text book - as my hobby is - I found the following question:
Find the equation of the tangent to the curve $y=\frac{5}{12}x^3-\frac{13}{9}x$ at the point P at which $x = x_0$. Show that the x-coordinate of the point Q where this tangent meets the curve again is $-2x_0$ and find the values of $x_0$ for which the tangent at P is the normal at Q.
I worked out the first part OK. I found the tangent to be $y=(\frac{5}{4}x_0^2-\frac{13}{9})x - \frac{5}{6}(x_0^3)$
I thought for the second part I need to solve this equation:
$(\frac{5}{4}x_0^2-\frac{13}{9})x - \frac{5}{6}(x_0^3)=\frac{5}{12}x-\frac{13}{9}x$
But I cannot solve it to get the answer in the text book, which is $\pm\frac{1}{3}\sqrt5,\pm\frac{2}{3}\sqrt2$
My workings took me to:
$x=\frac{45x^3-156x+90x_0^3}{135x_0^2-156}$
but this seems a dead end.
Your first part of the solution is correct. For the second part, you need to equate $y$ value of the curve and the tangent line.
$ \displaystyle \left(\frac{5}{4}x_0^2-\frac{13}{9}\right)x - \frac{5}{6}x_0^3=\frac{5}{12}x^3 -\frac{13}{9}x$
$ \displaystyle \frac{5}{4}x_0^2 \ x - \frac{5}{6}x_0^3=\frac{5}{12}x^3$
$ \displaystyle 3 x_0^2 \ x - 2 x_0^3 = x^3$
$x^3 - 3 x_0^2 \ x + 2 x_0^3 = 0$
$(x- x_0)^2 (x + 2x_0) = 0$
and hence $x = - 2 x_0$ is the x-value of intersection point $Q$.
Now for the third part, find slope of the tangent at $Q$, just like you did for first part.
Product of the slopes of tangents at $P$ and $Q$ should be $-1$ as the tangent at $P$ is normal to the curve at $Q$. Find at what value of $x_0$, that happens.